5. Noetherian Rings Since I Have an Extra Day, I Decided to Go Back and Prove Some of the Basic Properties of Noetherian Rings That I Haven’T Already Proven

26 MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA

5. Noetherian rings Since I have an extra day, I decided to go back and prove some of the basic properties of Noetherian rings that I haven’t already proven. 5.1. Hilbert basis theorem. Recall that R is called a Noetherian ring iﬀthe ideal of R satisfy the ACC. We saw that this was equivalent to saying that every ideal α R has a ﬁnite set of generators a , ,a ⊆ 1 ··· n which means that a is the set of all R-linear combinations of the ai. Theorem 5.1 (Hilbert Basis Theorem). If R is Noetherian then so is R[X].

Proof. Suppose that I is an ideal in R[X]. For each n 0 let an be the set consisting of 0 and all leading coeﬃcients of polynomials≥ in I of degree n. This can also be described as the set of all a R so that I contains an element of the form ∈ n n 1 n 2 aX + a X − + a X − + + a 1 2 ··· n Claim 1. an is an ideal of R or is equal to R. Proof: To show that an is an ideal, choose two elements a, b an. This is equivalent to saying that I contains two polynomials of the∈ form f(X)=aXn + lower terms g(X)=bXn + lower terms Suppose that r, s R. Then rf + sg I. But ∈ ∈ rf + sg =(ra + sb)Xn + lower terms

Therefore, ra + sb an. So, an is an ideal in R (or an = R). Claim 2. a a∈ . m ⊆ n+1 This statement is obvious: If a an then I contains a polynomial f(X)=aXn+ . Since X R[X],∈I also contains Xf(X)=aXn+1+ ··· ∈ . So, a an+1. ···Since R ∈is Noetherian, the ascending sequence a a a 0 ⊆ 1 ⊆ 2 ⊆··· stops at some point N and we get a = a = . N N+1 ··· For i =0, ,N let aij R be a ﬁnite set of generators for the ideal a . Let f ···I be a polynomial∈ of degree i with leading coeﬃcient a . i ij ∈ ij Claim 3. The polynomials fij for i =0, ,N generate I as an ideal over R[X]. ··· I proved this by induction on n: If f(X) is a polynomial in I of degree n then f(X) is equal to a polynomial of degree less than n plus some R[X]-linear combination of the polynomials fij. To prove this I considered two cases. MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 27

Case 1. Suppose first that n>N. In that case f(X)=aXn + . ··· Since a an = aN there are elements rj R so that a = rjaNj. ∈ N ∈ This means that rjfNj(X)=aX + . So, ··· ! n N ! f(X) rjX − fNj(X) − is an element of I of degree Noetherian ring. Lemma 5.2. Any quotient of a Noetherian ring is Noetherian. And then we have the following consequence of this lemma and the Hilbert basis theorem. Corollary 5.3. If R is Noetherian, then any finitely generated ring over R is also finitely generated. For example, any finitely generated ring over a field is Noetherian. 5.2. Noetherian modules. The next basic property was the third equivalent definition of a Noetherian ring: Theorem 5.4. A ring R is Noetherian if and only if every submodule of a finitely generated R module is finitely generated. I pointed out that one direction ( ) is obvious since R is a finitely generated module over itself and the⇐ proper submodules of R are the ideals. To prove the converse I needed some definitions. Definition 5.5. Suppose that R is any ring. Then an R-module M is called Noetherian if the submodules of M satisfy the ACC. This is clearly equivalent to saying that every submodule of M is finitely generated. Using this definition, a ring R is Noetherian if and only if R is a Noetherian R-module. From this we want to conclude that every f.g. R-module is Noetherian. But every f.g. R-module is the quotient of a f.g. free module Rn. Therefore, the theorem follows from the following two lemmas. 28 MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA

Lemma 5.6. If M is a Noetherian R-module then so is every quotient module of M. This is obvious because the submodules of any quotient M/N corre- spond to the submodules of M containing N. Lemma 5.7. If M,N are Noetherian R-modules then so is M N. ⊕ I proved the following generalization of this lemma. And I pointed out that since the proposition proves the lemma and the lemma proves the theorem we will be done. Proposition 5.8. Suppose that 0 K M π N 0 is a short exact sequence of R-modules. Then→ tfae. → −→ → (1) M is Noetherian. (2) K and N are both Noetherian. Proof. It is clear that the ﬁrst statement implies the second. So, suppose that the second statement is true. Let L1 L2 be an ascending sequence of submodules of M. Then we⊆ want⊆ to··· show that the sequence stops. Let Ai = Li K. This is an increasing sequence of submodules of K. So, for large∩ enough N, we have A = A = . N N+1 ··· Let Bi = π(Li) be the image of Li in N. Then, because N is Noe- therian, there is a large number N " which we can take to be equal to N so that B = B = . N ! N !+1 ··· Now we claim that LN = LN+1 = . This is a special case of the 5-lemma applied to the following diagram.··· - - - - 0 AN LN BN 0

= = ? ? ? - - - - 0 AN+1 LN+1 BN+1 0 Since the rows are exact and the diagram commutes, the mapping L # L is an isomorphism. I.e., L = L . N → N+1 N N+1 ! MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 29

5.3. Associated primes. The idea is to obtain something similar to eI the unique factorization theorem for integers: n = pi where we replace pi with prime ideals in a Noetherian ring R and we replace n with a f.g. R-module M. For example, 12 = 223# corresponds to the statement that the prime ideals (2), (3) Z are associated to the ⊆ Z-module M = Z/(12). There are three ways to associate prime ideals to modules. Two are easy and we also use a third method which has better properties. The first way is to look at the cyclic submodules. For example, M = Z/(12) contains submodules isomorphic to Z/(2) and Z/(3). The other simple method is to look at the quotient modules. There is an epimorphism Z/(12) " Z/(p) only for the primes p =2, 3. However, there tend to be more quotients than submodules. For example, Z/(p) is a quotient of M = Z for every prime p. But the only prime ideal p Z so that ⊂ Z/p is embedded in Z is p = 0. 5.3.1. annihilators and associated primes. I will always assume that R and M are both Noetherian. Definition 5.9. If x =0 M then the annihilator ann(x) is the set of all a R so that ax+ = 0.∈ It is easy to see that this is an ideal in R. ∈ Definition 5.10. For any R-module M, the associated primes are the prime ideals p which occur as annihilators of nonzero elements of M. This is equivalent to the statement that there is a monomorphism R/p # M → The set of associated ideals is denoted ass(M). So, ass(M)= p R prime p = ann(x) for some x M { ⊂ | ∈ } Example 5.11. If p R is prime then ass(R/p)= p . ⊂ { } Proof. Clearly, p is a prime associated to R/p. Suppose that q is an- other associated prime. Then q = ann(x) for some x = y + p in R/p. But then q = ann(x)= a R ax =0 R/p { ∈ | ∈ } = a R ay p = p { ∈ | ∈ } ! The next theorem is that the union of the associated primes is the set of zero divisors. Definition 5.12. a R is called a zero divisor of M if ax = 0 for some nonzero x M.∈ In other words, the set of zero divisors of M is equal to the union∈ of all ann(x) for all nonzero x M. ∈ 30 MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA

I pointed out that a R is not a zero divisor of M if and only if multiplication by a does∈ not kill any nonzero element of M. I.e., a = a : M M · M → is a monomorphism. In that case a is called M-regular. Theorem 5.13. The union of the associated primes of M is equal to the set of zero divisors of M: p = zero divisors of M { } p ann(M) ∈ $ First we need a lemma: Lemma 5.14. Consider the set of all ideals I R so that I = ann(x) for some x R. Then the maximal elements of⊂ this set are prime. ∈ Proof. Suppose that I = ann(x) is maximal but not prime. Then there exist a, b R so that ab I but a, b / I. Since b/I, bx = 0. But abx = 0. So,∈ a ann(bx) and∈ a/I ann(∈x). This implies∈ that+ ann(bx) ∈ ∈ is strictly larger than ann(x) which is a contradiction. ! 5.3.2. localization and support. There are two other methods to associate primes to R-modules. The easy way is to take the set of all p so that M/pM = 0. However, localization is a better method even though it is more complicated.+ 1 Recall that, for any multiplicative subset S of R, S− M is the R module given by: 1 S− M = x/s x M, s S / { | ∈ ∈ } ∼ where x/s y/t iff rtx = rsy for some r S. If p is a prime ideal ∼ ∈ 1 then the localization Mp of M at p is defined to be Mp = S− M where S is the complement of p in R. Definition 5.15. The support of M is defined to be the set of all primes p so that M = 0: p + supp(M) := p R prime M =0 { ⊂ | p + } Example 5.16. supp(R/I)= p R prime I p . { ⊂ | ⊆ } This follows from the elementary fact that 1 S− (R/I)=0 S I = ⇐⇒ ∩ + ∅ The advantage of the localization functor M Mp is that it is exact while M M/pM is not exact. /→ /→ Proposition 5.17. Suppose that S R is a multiplicative set. Then 1 ⊂ the functor M S− M is exact. /→ MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 31

ψ φ Proof. Suppose that N M L is exact, i.e., ker φ = im ψ. Then we want to show that −→ −→

1 ψ 1 φ 1 S− N S− M S− L −→ −→ 1 is exact. The composition is certainly 0. So, suppose that x/s S− M ∈ is in the kernel of φ. Then φ(x)/s = 0. So, t S s.t. tφ(x)= 0=φ(tx). This implies that tx = ψ(y) for∃ some∈ y N. Then ∈ x/s = tx/ts = ψ(y/ts). So, ker φ = im ψ. ! An immediate consequence of the exactness of localization is the following. Corollary 5.18. If 0 N M M/N 0 is a short exact sequence of R-modules then→ → → → supp(M) = supp(N) supp(M/N) ∪ Proof. For any prime p we get a short exact sequence 0 N M (M/N) 0 → p → p → p → p supp(M) Mp =0 either Np = 0 or (M/N)p =0 p ∈ supp(N) ⇐⇒supp(M/N+ ). ⇐⇒ + + ⇐⇒ ∈ ∪ ! The main result relating associated primes and the support is the following. Theorem 5.19. Suppose that R and M are Noetherian. Then ass(M) supp(M) ⊆ Furthermore, the minimal elements of supp(M) are associated primes. The minimal elements of supp(M) are called the minimal (or iso- lated) associated primes. Proof. Suppose that p is an associated prime. Then we have a short exact sequence 0 R/p M X 0 → → → → By exactness of localization, we get an exact sequence 0 (R/p) M X 0 → p → p → p → But (R/p)p = 0 by Example 5.16. So, Mp = 0. So p is in the support of M. + + Now suppose that p supp(M) is minimal. Then Mp = 0. So it has a nonzero element x/s∈ . This being nonzero means that+ tx = 0 for + 32 MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA all t S (where S is the complement of p in R). Consequently, the annihilator∈ of x/s is disjoint from S. So ann(x) ann(x/s) p ⊆ ⊆ Let q = ann(x/s) be maximal. Then q ass(M ) by Lemma 5.14. ∈ p Since R is Noetherian, q =(a1, ,an). This means that there are elements t , ,t S so that t a···x = 0. This implies that 1 ··· n ∈ i i q ann( t x) ann( t x/s) ⊆ i ⊆ i But ann( tix/s)=q by maximality# of q.# So,

# q = ann( tix) ass(M) supp(M) ∈ ⊆ which implies that q = p #ass(M). ∈ ! 5.3.3. intersection of associated primes. One consequence of the theorem is that the intersection of the associated primes is the same as the intersection of the primes in the support. Corollary 5.20. p = p = rad(ann(M)) p ass(M) p supp(M) ∈ % ∈ % To prove this we need the following lemma. Lemma 5.21. The radical of any ideal I is the intersection of all prime ideals containing it: rad I = p I p prime ⊆ % For this we need a description of the embedding 1 Spec(S− R) # Spec(R) → where Spec(R) is the set of prime ideals in R. This embedding sends 1 any ideal a in S− R to the ideal I = a R := a R a/1 a | { ∈ | ∈ } This is clearly an ideal in R which is disjoint from S (otherwise a 1 contains a unit s/1). Also a prime implies S− R/a is a domain which imlies that a R is prime since it is the kernel of the composition | 1 1 R S− R S− R/a. → → 1 (a I = a R describes an embedding since a = S− I.) /→ | MATH 101B: ALGEBRA II PART B: COMMUTATIVE ALGEBRA 33

Proof of the lemma. ( ) is clear since I p rad I p. ( ) If a/rad(I) then⊆ the multiplicative⊆ ⇒ set S =⊆ 1, a, a2, is ⊇ ∈ 1 { ···} disjoint from I. This implies that the ring T − (R/I) is nonzero (T is the image of S in R/I). So, it has a maximal (and thus prime) ideal m. Then P = m R/I is a prime in R/I disjoint from T . Then P = p/I for some prime ideal| p containing I and disjoint from S. This implies that a/p. So, a/ p. ∈ ∈ ! Proof of corollary. Since& the minimal associated primes are the same as the minimal supporting primes, the two intersections are equal. ( ) ann(M) ann(x) for all x M. In particular, ann(M) is contained⊇ in every⊆ associated prime ∈p. But then rad(ann(M)) is also contained in each p. ( ) Suppose that a/rad(ann(M)). Then, by the lemma, there is a prime⊆ ideal p containing∈ ann(M) so that a/p. But p ann(M) M = 0. So, p is a supporting prime. So, a/∈ p. ⊆ ⇒ p + ∈ ! 5.3.4. primes associated to extensions. & Theorem 5.22. Suppose that 0 N M M/N 0 is a short exact sequence of R-modules. Then→ → → → ass(N) ass(M) ass(N) ass(M/N) ⊆ ⊆ ∪ Proof. The ﬁrst inclusion is obvious: If R/p embeds in N then it embeds in M. So, suppose that p is associated to M. Then p = ann(x) for some x M. There are two cases. Either Rx N = 0 or Rx N = 0. ∈ ∩ ∩ + If Rx N = 0 then R/p ∼= Rx embeds in M/N making p an associated prime∩ of M/N. If Rx N contains y = ax = 0 in particular, a/ann(x)=p. The annihilator∩ of y is the set of all+ b R so that bax =∈ 0. But this implies ba p. Since a/p we must have∈b p. So, ann(y)=p ass(N). ∈ ∈ ∈ ∈ The theorem holds in both cases. ! When the short exact sequence splits, this theorem implies that ass(M) = ass(N) ass(M/N) ∪ By induction this implies the following.

Corollary 5.23. ass( Mi)= ass(Mi). ' (